我正在使用机器人框架和请求库测试REst API API的JSON响应采用表格的形式。 我正在使用机器人框架来测试这个API,我想将这个表拆分成字典,以便我可以使用robot framework关键字来测试响应。
以下是我的JSON响应示例:
[
{
locale: "fr_FR",
day: "2015-12-01",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
{
locale: "fr_FR",
day: "2015-12-02",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
[
理想情况下,我想以字典的形式提取整个响应,以便我可以遍历响应并断言键和值。
所以我不会将字典嵌入到表中,而只会使用字典:
{
locale: "fr_FR",
day: "2015-12-01",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
{
locale: "fr_FR",
day: "2015-12-02",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
我尝试过收藏和请求库但是我收到了错误:
${JSON}= To JSON ${resp.content}
${DICT}= Convert To List ${JSON}
Log ${DICT}
:FOR ${KEY} IN locale day displayPricePerPerson displayTotalPrice promotion minLos
\ Run Keyword And Continue On Failure List Should Contain Value ${JSON} ${KEY}
Error : ValueError: dictionary update sequence element #0 has length 6; 2 is required
答案 0 :(得分:1)
you have dictionaries in your list already. though badly formed.they are missing quotes around the keys
lt = [
{
"locale": "fr_FR",
"day": "2015-12-01",
"displayPricePerPerson": 9800,
"displayTotalPrice": 9800,
"promotion": False,
"minLos": 1
},
{
"locale": "fr_FR",
"day": "2015-12-02",
"displayPricePerPerson": 9800,
"displayTotalPrice": 9800,
"promotion": False,
"minLos": 1
},
]
for el in lt:
print(type(el))
<class 'dict'>
<class 'dict'>
d1 = lt[0]
d2 =lt[1]
print(d1.items())
dict_items([('day', '2015-12-01'), ('locale', 'fr_FR'), ('displayPricePerPerson', 9800), ('minLos', 1), ('promotion', False), ('displayTotalPrice', 9800)])
print(d2.items())
dict_items([('day', '2015-12-02'), ('locale', 'fr_FR'), ('displayPricePerPerson', 9800), ('minLos', 1), ('promotion', False), ('displayTotalPrice', 9800)])
to convert to list of dicts this:
lt2 = '''[
{
locale: "fr_FR",
day: "2015-12-01",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
{
locale: "fr_FR",
day: "2015-12-02",
displayPricePerPerson: 9800,
displayTotalPrice: 9800,
promotion: false,
minLos: 1
},
]
'''
def convert_ECMA_Javascript(st):
# convert badly formed strings to json format
import json
result = re.sub(r'(\w+):',r'"\1":',st)
result= re.sub(r'false',r'"False"',result)
result= re.sub(r'true',r'"True"',result)
result= re.sub(r'\[|\]',r'',result)
result= re.sub(r'\b(\d+)\b(?!"|-)',r'"\1"',result)
result= re.sub(r'\n|\t',r'',result)
li = re.findall(r'{.*?}', result)
result = []
for s in li:
result.append(json.loads(s))
return result
pp(convert_ECMA_Javascript(lt2))
[{'day': '2015-12-01',
'displayPricePerPerson': '9800',
'displayTotalPrice': '9800',
'locale': 'fr_FR',
'minLos': '1',
'promotion': 'False'},
{'day': '2015-12-02',
'displayPricePerPerson': '9800',
'displayTotalPrice': '9800',
'locale': 'fr_FR',
'minLos': '1',
'promotion': 'False'}]
for el in convert_ECMA_Javascript(lt2):
for k,v in el.items():
print(k,":", v)
day : 2015-12-01
displayPricePerPerson : 9800
minLos : 1
promotion : False
displayTotalPrice : 9800
locale : fr_FR
day : 2015-12-02
displayPricePerPerson : 9800
minLos : 1
promotion : False
displayTotalPrice : 9800
locale : fr_FR
答案 1 :(得分:1)
假设实际数据是有效的JSON(问题中的数据不是),一旦调用To JSON
,返回的数据就是可以迭代的字典列表。您应该能够在没有任何进一步转换的情况下循环它们。
例如:
${JSON}= To JSON ${resp.content}
:FOR ${item} IN @{JSON}
\ log locale: ${item["locale"]}
以上内容将为JSON数据中的每个字典记录一行。